Functions from $\mathbb{R}$ to $\mathbb{R}$ and axiom of choice I am curious about a "contradiction" in the definition of functions that has long been unsolved in my head.
If we define a function $f$ from a set $A$ to $B$ as an object such thatf:
" $\forall a \in A. \exists ! b \in B. b = f(a)$",
then such an object could then be equally defined as $\prod_{a \in A} f(a)$, that is, a member of $\prod_A B$.
For $A=\mathbb{R}$, such an set can only exist if the axiom of choice holds. How does one get round this contradiction ?
 A: The axiom of choice is not needed in order to construct a function from $\Bbb R$ to $\Bbb R$: for instance, every polynomial with concretely specified coefficients is a function from $\Bbb R$ to $\Bbb R$ whose existence does not depend on the axiom of choice, the most obvious examples being the constant functions and the identity function.
The axiom of choice is equivalent to the statement that all Cartesian products of non-empty sets are non-empty; this does not mean that we can never prove that a particular Cartesian product of non-empty sets is non-empty without using the axiom of choice.
A: The statement of the axiom of choice is as follows:
Let $\{A_i\}_{i \in I}$ be an indexed collection of sets. Then, there exists a function $f: I \to \bigcup_{i \in I} A_i$ such that:
$$\forall i \in I: f(i) \in A_i$$
If what you say is true, then we are defining a function using a function. This is circular and certainly not correct.
Now, you bring up the fact that "such an object could be equally defined as $\prod_{a \in A} f(a)$". What does that even mean? When you're taking the product of the images of the elements of $A$ under $f$, what exactly are you doing? Remember, products are functions too so that doesn't seem to be equivalent at all.
I should also mention that $\prod_{A} B$ has no meaning. You should clarify what you mean by that.
The definition of a function is very simple. Let $A$ and $B$ be sets. Then, a function $f: A \to B$ is a set $f \subset A \times B$ that is defined by the following proposition:
$$\forall x \in A: \exists ! y \in B: (x,y) \in f$$
This is just saying that we can construct a relation such that each element of $A$ is assigned exactly one element of $B$. This doesn't require the axiom of choice.
